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Difference between revisions of "1985 AHSME Problems/Problem 29"

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<math> \mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \  } 17865 \qquad \mathrm{(D) \  } 17874 \qquad \mathrm{(E) \  }19851 </math>
 
<math> \mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \  } 17865 \qquad \mathrm{(D) \  } 17874 \qquad \mathrm{(E) \  }19851 </math>
  
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==Solution==
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==Solution 1==
 
By the formula for the sum of a geometric series, <cmath>\begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \\ &= \frac{8\left(10^{1985}-1\right)}{10-1} \\ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*}</cmath> and similarly <cmath>b = \frac{5\left(10^{1985}-1\right)}{9},</cmath> so <cmath>\begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \\ &= \frac{40\left(10^{1985}-1\right)^2}{9} \\ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \\ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}</cmath>
 
By the formula for the sum of a geometric series, <cmath>\begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \\ &= \frac{8\left(10^{1985}-1\right)}{10-1} \\ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*}</cmath> and similarly <cmath>b = \frac{5\left(10^{1985}-1\right)}{9},</cmath> so <cmath>\begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \\ &= \frac{40\left(10^{1985}-1\right)^2}{9} \\ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \\ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}</cmath>
  
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Using long division, and noting that <math>80 = 8 \cdot 9 + 8</math> and <math>81 = 9 \cdot 9</math>, it follows that <cmath>\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,</cmath> with <math>1984</math> ones, <math>1</math> zero, then <math>1984</math> eights, <math>1</math> nine, and a final zero. Lastly, using long multiplication and noting that <math>9 \cdot 4 = 36</math>, <math>8 \cdot 4 = 32</math>, and <math>8 \cdot 4 + 3 = 35</math>, we obtain <cmath>\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,</cmath> where there are <math>1984</math> fours, <math>1</math> three, <math>1984</math> fives, <math>1</math> six, and a final zero, so the sum of the digits is <cmath>\begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{\text{(C)} \ 17865}.\end{align*}</cmath>
 
Using long division, and noting that <math>80 = 8 \cdot 9 + 8</math> and <math>81 = 9 \cdot 9</math>, it follows that <cmath>\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,</cmath> with <math>1984</math> ones, <math>1</math> zero, then <math>1984</math> eights, <math>1</math> nine, and a final zero. Lastly, using long multiplication and noting that <math>9 \cdot 4 = 36</math>, <math>8 \cdot 4 = 32</math>, and <math>8 \cdot 4 + 3 = 35</math>, we obtain <cmath>\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,</cmath> where there are <math>1984</math> fours, <math>1</math> three, <math>1984</math> fives, <math>1</math> six, and a final zero, so the sum of the digits is <cmath>\begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{\text{(C)} \ 17865}.\end{align*}</cmath>
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==Solution 2==
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Factoring out the 8 and the 5, we get <math>9ab = 10\cdot4\cdot9\cdot111 \dotsb 1 \ cdot 111\dotsb1,</math> where there are 1985 1s in both numbers
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 +
This is equal to <cmath>10 \cdot 4 \cdot (1000\dotsb 0 - 1) \cdot 111\dotsb 1</cmath> where there is an initial 1 followed by 1985 0s and the second number has 1985 1s.
 +
 +
Multiplying then subtracting gives us <math>4 \cdot 10 \cdot 111\dotsb 10888 \dotsb 89,</math> where there are 1984 1s and 1984 8s.
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 +
Multiplying one last time, then adding the 0 at the and gives us <math>44 \dotsb 4355 \dotsb 560,</math> where there are <math>1984</math> fours, <math>1</math> three, <math>1984</math> fives, <math>1</math> six, and a final zero, so the sum of the digits is <math>1984(4+5)+6+3 = 1985 \cdot 9 = 17865</math>
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=28|num-a=30}}
 
{{AHSME box|year=1985|num-b=28|num-a=30}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:51, 26 March 2025

Problem

In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$?

$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qquad \mathrm{(D) \ } 17874 \qquad \mathrm{(E) \ }19851$

Solution 1

By the formula for the sum of a geometric series, \begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \\ &= \frac{8\left(10^{1985}-1\right)}{10-1} \\ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*} and similarly \[b = \frac{5\left(10^{1985}-1\right)}{9},\] so \begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \\ &= \frac{40\left(10^{1985}-1\right)^2}{9} \\ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \\ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}

We now compute the decimal expansion of this expression. Firstly, $10^{3971} = 100 \dotsb 0$, with $1$ one and $3971$ zeroes, and $2 \cdot 10^{1986} = 200 \dotsb 0$, with $1$ two and $1986$ zeroes. Subtracting therefore gives \[10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,\] where there are $3971-1986-1 = 1984$ nines followed by $1$ eight and then $1986$ zeroes. Adding $10$ transforms this to $99 \dotsb 9800 \dotsb 010$, now with $1984$ nines followed by $1$ eight, $1984$ zeroes, $1$ one, and a final zero.

Using long division, and noting that $80 = 8 \cdot 9 + 8$ and $81 = 9 \cdot 9$, it follows that \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,\] with $1984$ ones, $1$ zero, then $1984$ eights, $1$ nine, and a final zero. Lastly, using long multiplication and noting that $9 \cdot 4 = 36$, $8 \cdot 4 = 32$, and $8 \cdot 4 + 3 = 35$, we obtain \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,\] where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is \begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{\text{(C)} \ 17865}.\end{align*}

Solution 2

Factoring out the 8 and the 5, we get $9ab = 10\cdot4\cdot9\cdot111 \dotsb 1 \ cdot 111\dotsb1,$ where there are 1985 1s in both numbers

This is equal to \[10 \cdot 4 \cdot (1000\dotsb 0 - 1) \cdot 111\dotsb 1\] where there is an initial 1 followed by 1985 0s and the second number has 1985 1s.

Multiplying then subtracting gives us $4 \cdot 10 \cdot 111\dotsb 10888 \dotsb 89,$ where there are 1984 1s and 1984 8s.

Multiplying one last time, then adding the 0 at the and gives us $44 \dotsb 4355 \dotsb 560,$ where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is $1984(4+5)+6+3 = 1985 \cdot 9 = 17865$

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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